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Chapter 5: Problem 13

A stone of mass \(0.3 \mathrm{~kg}\) attached to a \(1.5 \mathrm{~m}\) long string is whirled around in a horizontal circle at a speed of \(6 \mathrm{~m} / \mathrm{s}\). The tension in the string is: (a) \(10 \mathrm{~N}\) (b) \(20 \mathrm{~N}\) (c) \(7.2 \mathrm{~N}\) (d) none of these

Short Answer

Expert verified
The tension in the string is 7.2 N (option c).

Step by step solution

01

Understand the Problem

We need to find the tension in the string when a stone of mass \(0.3 \mathrm{~kg}\) is being whirled in a horizontal circle at a speed of \(6 \mathrm{~m/s}\). The length of the string is \(1.5 \mathrm{~m}\). The stone is undergoing circular motion, so tension provides the centripetal force.
02

Write the Equation for Centripetal Force

The formula for centripetal force \(F_c\) required to keep an object moving in a circle of radius \(r\) with speed \(v\) is given by: \[ F_c = \frac{m v^2}{r} \]where \(m\) is the mass of the object, \(v\) is the speed, and \(r\) is the radius of the circle.
03

Substitute the Known Values

Plug in the values into the centripetal force equation: - \(m = 0.3 \mathrm{~kg}\) (mass of the stone)- \(v = 6 \mathrm{~m/s}\) (speed of the stone)- \(r = 1.5 \mathrm{~m}\) (radius, same as the string's length)\[ F_c = \frac{0.3 \times 6^2}{1.5} \]
04

Calculate the Expression

Calculate the expression step by step:- First, calculate \(v^2 = 6^2 = 36\)- Then, multiply by the mass: \(0.3 \times 36 = 10.8\)- Finally, divide by the radius: \(\frac{10.8}{1.5} = 7.2 \text{ N}\)
05

Compare with Given Options

The calculated tension in the string is \(7.2 \mathrm{~N}\), which matches option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Motion
Circular motion occurs when an object moves in a path that forms a circle. This motion is characterized by a continuous change in direction, which implies that the object is constantly accelerating, even if its speed remains constant. In the case of the stone tied to a string, the stone travels in a pure circular path because it is being whirled around.
  • The stone experiences a force directed towards the center of the circle. This force is known as centripetal force.
  • Circular motion is common in many physical applications such as satellite orbits, spinning wheels, and the rotation of gears.
To maintain this motion without changing the speed, a constant inward force must be applied. This is where the tension in the string plays a significant role.
Tension in String
The tension in the string acts as the centripetal force required to keep the stone moving in its circular path. Tension here refers to the force transmitted through the string, pulling the massive stone towards the center of the circle.
  • The tension force directly balances out the stone's inertia that tries to move it in a straight line.
  • It can be calculated using the formula for centripetal force.
  • Higher tension results from either an increase in mass, speed, or a decrease in the radius of the circle.
In problems involving tension and circular motion, understanding how these physical entities interact is crucial. The tension is the invisible hand that controls the stone鈥檚 path.
Physics Problem Solving
Solving physics problems efficiently involves a logical and structured approach. When tackling problems like determining the tension in a string during circular motion, follow these general steps:
  • Understand the problem: Identify what is being asked and the physical principles involved, such as circular motion for tension problems.
  • Formulate equations: Use relevant equations, like the centripetal force equation (\[ F_c = \frac{m v^2}{r} \]), to express known quantities.
  • Substitute values: Carefully substitute known values into equations ensuring correct units for mass, speed, and radius.
  • Calculate results: Solve the equation step by step to find the desired value, such as the tension, without skipping steps.
This disciplined approach helps build confidence and accuracy in problem solving, especially in physics.
Centripetal Acceleration
Centripetal acceleration is the rate of change of velocity that keeps an object moving along a circular path. It points towards the center of the circle, much like the centripetal force.
  • It is given by the expression: \[ a_c = \frac{v^2}{r} \]where \( a_c \) is the centripetal acceleration.
  • This acceleration does not alter the speed of the object, only its direction.
  • Both centripetal force and centripetal acceleration are crucial for maintaining circular motion, resisting the inertia that pushes the object outward.
In our stone-and-string example, the centripetal acceleration is the radial acceleration that the stone experiences. Together with tension, it ensures the stone keeps on its circular trajectory without flying off, clarifying how subtle physics concepts create balanced forces in motion.

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Most popular questions from this chapter

A tube of iength \(L\) is filled completely with an incompressible liquid of mass \(M\) and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity \(\omega\). The force exerted by the liquid at the other end is:

A solid body rotates about a stationary axis so that its angular velocity depends on the rotational angle \(\phi\) as \(\omega=\omega_{0}-k \phi\) where \(\omega_{0}\) and \(k\) are positive constants. At the moment \(t=0, \phi=0\), the time dependence of rotation angle is: (a) \(k \omega_{0} e^{-k t}\) (b) \(\frac{\omega_{0}}{k} e^{-k t}\) (c) \(\frac{\omega_{0}}{k}\left(1-e^{-k t}\right)\) (d) \(\frac{k}{\omega_{0}}\left(e^{-k t}-1\right)\)

Mark correct option or options from the following : (a) In the case of circular motion of a particle, centripetal force may be balanced by centrifugal force (b) In the non-inertial reference frame centrifugal force is real force (c) In the inertial reference frame, centrifugal force is real force (d) Centrifugal force is always pseudo force

A rod OA rotates about a horizontal axis through \(O\) with a constant anticlockwise velocity \(\omega=3\) rad/sec. As it passes the position \(\theta=0\) a small block of mass \(m\) is placed on it at a radial distance \(r=450 \mathrm{~mm}\). If the block is observed to slip at \(\theta=50^{\circ}\), the coefficient of static friction between the block and the rod is: (Given that \(\sin 50^{\circ}=0.766, \cos 50^{\circ}=0.64\) ) (a) \(0.2\) (b) \(0.55\) (c) \(0.8\) (d) 1

The angular displacement of the rod is defined as \(\theta=\frac{3}{20} t^{2}\) where \(\theta\) is in radian and \(t\) is in second. The collar \(B\) slides along the rod in such a way that its distance from \(O\) is, \(r=0.9-0.12 t^{2}\) where \(r\) is in metre and \(t\) is in second. The velocity of collar at \(\theta=30^{\circ}\) is: (a) \(0.45 \mathrm{~m} / \mathrm{s}\) (b) \(0.48 \mathrm{~m} / \mathrm{s}\) (c) \(0.52 \mathrm{~m} / \mathrm{s}\) (d) \(0.27 \mathrm{~m} / \mathrm{s}\)
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